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Number of balanced partitions of n: the largest part equals the number of parts.
115

%I #106 Oct 16 2024 09:55:06

%S 1,0,1,1,1,1,3,2,4,4,6,7,11,11,16,19,25,29,40,45,60,70,89,105,134,156,

%T 196,232,285,336,414,485,591,696,839,987,1187,1389,1661,1946,2311,

%U 2702,3201,3731,4400,5126,6018,6997,8195,9502,11093,12849,14949,17281,20062

%N Number of balanced partitions of n: the largest part equals the number of parts.

%C Useful in the creation of plane partitions with C3 or C3v symmetry.

%C The function T[m,a,b] used here gives the partitions of m whose Ferrers plot fits within an a X b box.

%C Central terms of triangle in A063995: a(n) = A063995(n,0). - _Reinhard Zumkeller_, Jul 24 2013

%C Sequence enumerates the collection of partitions of size n that are in the semigroup of Dyson rank=0, or balanced partitions, under the binary operation A*B = (a1,a2,...,a[k-1],k)*(b1,...,b[n-1,n) = (a1*b1,...,a1*n,a2*b1,...,a2*n,...,k*b1,...,k*n), where A is a partition with k parts and B is a partition with n parts, and A*B is a partition with k*n parts. Note that the rank of A*B is 0, as required. For example, the product of the rank 0 partitions (1,2,3) of 6 and (1,1,3) of 5 is the rank 0 partition (1,1,2,2,3,3,3,6,9) of 30. There is no rank zero partition of 2, as shown in the sequence. It can be seen that any element of the semigroup that partitions an odd prime p or a composite number of form 2p cannot be a product of smaller nontrivial partitions, whether in this semigroup or not. - _Richard Locke Peterson_, Jul 15 2018

%C The "multiplication" given above was noted earlier by _Franklin T. Adams-Watters_ in A122697. - _Richard Peterson_, Jul 19 2023

%C The Heinz numbers of these integer partitions are given by A106529. - _Gus Wiseman_, Mar 09 2019

%H Vaclav Kotesovec, <a href="/A047993/b047993.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from T. D. Noe)

%H Erich Friedman, <a href="/A047993/a047993.gif">Illustration of initial terms</a>

%F a(n) = p(n-1) - p(n-2) - p(n-5) + p(n-7) + ... + (-1)^k*(p(n-(3*k^2-k)/2) - p(n-(3*k^2+k)/2)) + ..., where p() is A000041(). E.g., A047993 a(20) = p(19) - p(18) - p(15) + p(13) + p(8) - p(5) = 490 - 385 - 176 + 101 + 22 - 7 = 45. - _Vladeta Jovovic_, Aug 04 2004

%F G.f.: Sum_{k>=1} (-1)^k * ( x^((3*k^2+k)/2) - x^((3*k^2-k)/2) ) ) / Product_{k>=1} (1-x^k). - _Vladeta Jovovic_, Aug 05 2004

%F a(n) ~ exp(Pi*sqrt(2*n/3))*Pi / (48*sqrt(2)*n^(3/2)) ~ p(n) * Pi / (4*sqrt(6*n)), where p(n) is the partition function A000041. - _Vaclav Kotesovec_, Oct 06 2016

%F G.f.: Sum_{n>=0} [2n,n]_q q^(2*n), where [m,n]_q are the q-binomial coefficients. - _Mamuka Jibladze_, Aug 12 2021

%F G.f.: Sum_{k>=1} x^(2*k-1) * Product_{j=1..k-1} (1-x^(k+j-1)/(1-x^j). - _Seiichi Manyama_, Jan 24 2022

%e From _Joerg Arndt_, Oct 08 2012: (Start)

%e a(12) = 7 because the partitions of 12 where the largest part equals the number of parts are

%e 2 + 3 + 3 + 4,

%e 2 + 2 + 4 + 4,

%e 1 + 3 + 4 + 4,

%e 1 + 2 + 2 + 2 + 5,

%e 1 + 1 + 2 + 3 + 5,

%e 1 + 1 + 1 + 4 + 5, and

%e 1 + 1 + 1 + 1 + 2 + 6.

%e (End)

%e From _Gus Wiseman_, Mar 09 2019: (Start)

%e The a(1) = 1 through a(13) = 11 integer partitions:

%e 1 21 22 311 321 322 332 333 4222 4322 4332 4333

%e 331 4211 4221 4321 4331 4422 4432

%e 4111 4311 4411 4421 4431 4441

%e 51111 52111 52211 52221 52222

%e 53111 53211 53221

%e 611111 54111 53311

%e 621111 54211

%e 55111

%e 622111

%e 631111

%e 7111111

%e (End)

%p A047993 := proc(n)

%p a := 0 ;

%p for p in combinat[partition](n) do

%p r := max(op(p))-nops(p) ;

%p if r = 0 then

%p a := a+1 ;

%p end if;

%p end do:

%p a ;

%p end proc:

%p seq(A047993(n),n=1..20) ; # _Emeric Deutsch_, Dec 11 2004

%t Table[ Count[Partitions[n], par_List/; First[par]===Length[par]], {n, 12}] or recur: Sum[T[n-(2m-1), m-1, m-1], {m, Ceiling[Sqrt[n]], Floor[(n+1)/2]}] with T[m_, a_, b_]/; b < a := T[m, b, a]; T[m_, a_, b_]/; m > a*b := 0; T[m_, a_, b_]/; (2m > a*b) := T[a*b-m, a, b]; T[m_, 1, b_] := If[b < m, 0, 1]; T[0, _, _] := 1; T[m_, a_, b_] := T[m, a, b]=Sum[T[m-a*i, a-1, b-i], {i, 0, Floor[m/a]}];

%t Table[Sum[ -(-1)^k*(p[n-(3*k^2-k)/2] - p[n-(3*k^2+k)/2]), {k, 1, Floor[(1+Sqrt[1+24*n])/6]}] /. p -> PartitionsP, {n, 1, 64}] (* _Wouter Meeussen_ *)

%t (* also *)

%t Table[Count[IntegerPartitions[n], q_ /; Max[q] == Length[q]], {n, 24}]

%t (* _Clark Kimberling_, Feb 13 2014 *)

%t nmax = 100; p = 1; s = 1; Do[p = Normal[Series[p*x^2*(1 - x^(2*k - 1))*(1 + x^k)/(1 - x^k), {x, 0, nmax}]]; s += p;, {k, 1, nmax + 1}]; Take[CoefficientList[s, x], nmax] (* _Vaclav Kotesovec_, Oct 16 2024 *)

%o (PARI)

%o N=66; q='q + O('q^N );

%o S=2+2*ceil(sqrt(N));

%o gf= sum(k=1, S, (-1)^k * ( q^((3*k^2+k)/2) - q^((3*k^2-k)/2) ) ) / prod(k=1,N, 1-q^k );

%o /* _Joerg Arndt_, Oct 08 2012 */

%o (PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, x^(2*k-1)*prod(j=1, k-1, (1-x^(k+j-1))/(1-x^j)))) \\ _Seiichi Manyama_, Jan 24 2022

%o (Haskell)

%o a047993 = flip a063995 0 -- _Reinhard Zumkeller_, Jul 24 2013

%Y Cf. A000700, A063995, A064173, A064174.

%Y Cf. A003114, A006141, A039900, A090858, A106529, A324516, A324518, A324520.

%Y Column 1 of A350879.

%K nice,nonn

%O 1,7

%A _Wouter Meeussen_